Everyone of us know that anything multiplied with zero (0) is always 0. But while solving problems in mathematics, especially for permutations and combinations, we always write the value of 0! as 1.

Have you ever thought why is 0 factorial 1?

## What is the Factorial of a Number?

The factorial of a number is the function that multiplies the number by every natural number below it till 1. Symbolically, factorial is represented as “$!$”.

So, $n$ factorial is the product of the first n natural numbers and is represented as $n!$.

Mathematically, $n! = n \times \left(n – 1\right) \times \left(n – 2\right) \times \left(n – 3\right) \times â€¦ \times 1$.

For example $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

It is quite clear from the above definition and example how to calculate the factorial of any whole number greater than or equal to one, but why is the value of $0! = 1$ despite the mathematical rule that anything multiplied by zero is equal to zero?

## The Definition of a Zero Factorial

The definition of a factorial is the product of all integers equal to or less in value to the original number. In other words, a factorial is the number of combinations possible with numbers less than or equal to that number.

Because zero has no numbers less than it but is still in and of itself a number, there is but one possible combination of how that data set can be arranged – **it cannot**.

This still counts as a way of arranging it, so by definition, a zero factorial is equal to $1$, just as $1!$ is equal to one because there is only a single possible arrangement of this data set.

## Proving Why Is 0 Factorial 1

Factorial of $n$ i.e., $n!$ is written as $n! = n \times \left(n âˆ’ 1\right) \times \left(n âˆ’ 2\right) \times â€¦ \times 1$.

The value of $n!$ from the above can be also written as $n \times \left(n âˆ’ 1\right)!$.

$â‡’n! = n \times \left(n âˆ’ 1\right)!$

Now, let’s consider the value of $n$ equal to $1$,

$â‡’1! = 1! \times \left(1 âˆ’ 1\right)!$

$â‡’1! =1! \times \left(0\right)!$

Now, here LHS = $1! = 1$.

And, RHS = $1! \times \left(0\right)! = 1 \times \left(0\right)! = \left(0\right)!$

Therefore, $\left(0\right)! = 1$.

## Conclusion

In mathematics, there are many such instances which seem incorrect at the first glance but are logically correct when we closely observe them. $0!$ is one such expression that seems to be 0 but actually is equal to 1.

## Practice Problems

- $0! = 0$
- True
- False

- $0! = 1!$
- True
- False

- Evaluate
- $ \frac {2!}{0!}$
- $ \frac {0!}{3!}$
- $7^{0!}$
- $ \left(0! \right)^{0!}$
- $5! \times 0! \times 0 \times 2!$

- If $x = 0!$, then $3!$ in terms of $x$ is?
- Find $x$, if $x! = 24 \times 0!$

## Recommended Reading

- Why Does Any Number Raised to Zero Power Always Give One?
- Learn Math With Python
- How to Avoid Silly Mistakes in Maths

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